DOC PREVIEW
GVSU EGR 345 - Advanced Control

This preview shows page 1-2 out of 6 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 6 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 6 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 6 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

Advanced ControlJoel Oostdyk, Joshua Hoekstra, Wade Bernreuter01/14/19Purpose:To investigate simple position proportional servo motor control.Theory:DC motors will apply a torque between the rotor and stator that is related to the applied voltage or current. When a voltage is applied the torque will cause the rotor to accelerate. For any voltage and load on the motor there will tend to be a final angular velocity due to friction and drag in the motor. And, for a given voltage the ratio between steady state torque and speed will be a straight line. The basic equivalent circuit model for the motor is shown below. We can develop equations for this model. This model must also include the rotational inertia of the rotor and any attached loads. On the left hand sideis the resistance of the motor and the 'back emf' dependent voltage source. On the right hand side the inertiacomponents are shown. The rotational inertia J1 is the motor rotor, and the second inertia is an attached disk. The model can now be considered as a complete system.Looking at this relationship we see a basic first order differential equation. We can measure motor properties using some basic measurements.In our modeled system, friction was considered to be a constant torque. Accounting for this, the differentialequation for the system then becomes:Rewriting the previous equation then yields:This differential equation can be applied to the circuit shown below in Figure 1.T TfKVsKRJTfKVsKR1KTf1RVs1RKKJK2R JTfJK VsJ RJ RKKTfRKVsFigure 1 – the op-amp circuit.Equipment:- Computer, PSE Dell 119- CADET, model 325-1401 ECL, serial 215526- Fluke, model 8010A, serial GVSU 2-92- PMC Power Supply #5, #9- Motor, 12 VDC, serial 40791205- Potentiometer #1, model A R500, Beckman Instruments- Wires- LM 675 op-amp, specs found at www.national.com- 2- 1K- resistors- 2.2K- resistorProcedure:1. Develop a method to measure coefficients of the differential equation for the system.2. Set up the circuit shown in Figure 1, replacing the 741 with a LM675 and using 16 volt supply sources rather than an 8 volt supply.3. Create a LabVIEW program to output desired voltages, and to read in voltage inputs.4. Connect a potentiometer to the drive shaft of the 12V DC motor, and wire the pot to the analog input ofthe DAC card.5. Vary the output voltages to the circuit and record the changing voltages over the pot.6. From the gathered data, determine the coefficients from the differential equation in step 1.Results:The voltages that represented position of the shaft with respect to time were recorded using LabVIEW. Using the relationship that an offset of .25V in LabVIEW corresponded to a -/2 radian shaft rotation the voltages gathered by LabVIEW were converted to radial positions. The graphs below are of this radial position vs. time.Trial 1:Figure 2 – Graph of radial position vs. time for trial 1.Trial 2:Figure 3 – Graph of radial position vs. time for trial 2.radial pos-45-40-35-30-25-20-15-10-5050 2 4 6 8 10 12time (s)radiansradial pos-202468100 1 2 3 4 5time (s)radiansTrial 3:Figure 4 – Graph of radial position vs. time for trial 3.From these position graphs the constant angular velocity that is reached is found to be the slope of the position curve. To summarize then:Motor Resistance: 19 -Angular Velocity Step on LabVIEWTrial 1 4.2 0 to 4 VoltsTrial 2 2.3 0 to 3 VoltsTrial 3 3.6 0 to 4 VoltsTable 1 – Angular velocity valuesDiscussionSince the angular velocity is constant the angular acceleration is zero which reduces the differential equation to:sfVKRTK Since we know 2 values for - and their corresponding Vs values a system of 2 equations with two unknowns can be set up and the values of K and Tf can be found.33.242.4DKDKSolving the system we get K=.526 Nm/A and D=-1.791.Since the motor resistance is known a value for the frictional torque can be found. This value is found to be: -.0496 Nm.The fact that this is such a small value is reflective of the shaft impingement. In other words, this minor value of the frictional torque relates to a higher efficiency of the motor.radial pos024681012141618200 1 2 3 4 5 6time(s)radiansConclusion:In this lab, a simple control circuit was set up in combination with LabVIEW to record characteristics of the circuit. A differential equation relating the input voltage and the rotational motion of the motor was then constructed. The data from LabVIEW was analyzed to find coefficients of the differential


View Full Document

GVSU EGR 345 - Advanced Control

Documents in this Course
Y Axis

Y Axis

2 pages

Load more
Download Advanced Control
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Advanced Control and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Advanced Control 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?